Editing Algebraic integer (section)

==Ring==
The sum, difference and product of two algebraic integers is an algebraic integer. In general their quotient is not. Thus the algebraic integers form a [[Ring (mathematics)|ring]].

This can be shown analogously to [[Algebraic_number#Field|the corresponding proof]] for [[algebraic number]]s, using the integers <math>\Z</math> instead of the rationals <math>\Q</math>.

One may also construct explicitly the monic polynomial involved, which is generally of higher [[degree of a polynomial|degree]] than those of the original algebraic integers, by taking [[resultant]]s and factoring. For example, if {{math|1=''x''<sup>2</sup> − ''x'' − 1 = 0}}, {{math|1=''y''<sup>3</sup> − ''y'' − 1 = 0}} and {{math|''z'' {{=}} ''xy''}}, then eliminating {{mvar|x}} and {{mvar|y}} from {{math|1=''z'' − ''xy'' = 0}} and the polynomials satisfied by {{mvar|x}} and {{mvar|y}} using the resultant gives {{math|1=''z''<sup>6</sup> − 3''z''<sup>4</sup> − 4''z''<sup>3</sup> + ''z''<sup>2</sup> + ''z'' − 1 = 0}}, which is irreducible, and is the monic equation satisfied by the product. (To see that the {{mvar|xy}} is a root of the {{mvar|x}}-resultant of {{math|''z'' − ''xy''}} and {{math|''x''<sup>2</sup> − ''x'' − 1}}, one might use the fact that the resultant is contained in the [[ideal (ring theory)|ideal]] generated by its two input polynomials.)

===Integral closure===
Every root of a monic polynomial whose coefficients are algebraic integers is itself an algebraic integer. In other words, the algebraic integers form a ring that is [[integrally closed domain|integrally closed]] in any of its extensions.

Again, the proof is analogous to [[Algebraic_number#Algebraic_closure|the corresponding proof]] for [[algebraic number]]s being [[algebraically closed field|algebraically closed]].